**Assembly Code Analysis**This section examines a simple assembly code snippet, detailing its operations:- **Line 0:** `LDA 8` - Load the content of the memory address 8 into the accumulator.- **Line 1:** `LDB 9` - Load the content of the memory address 9 into register B.- **Line 2 & 3:** `INP 8`, `INP 9` - Input values into memory addresses 8 and 9.- **Line 4:** `ADD` - Add the contents of the accumulator and register B.- **Line 5:** `STR 10` - Store the result at memory address 10.- **Line 6:** `PNT 10` - Print the value from memory address 10.- **Line 7:** `STP` - Stop the program.**Memory Content:**- **Address 8:** 10- **Address 9:** 20**Explanation:**This assembly code takes two inputs, adds them, and prints the result.

1. LDA 8 : "LDA" is the mnemonic for the opcode (load accumulator A) and "21" is the operand (an…

Answered: O LDA 8 1 LDB 9 2 INP 8 3 INP 9 4 ADD 5…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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MAer -0 and 03: SPI system shown in the following figure 1, if SS, = 1, SS, = 0, 5S, MISO = 1, state the data flow in such a case. %3D %3D = 1 SCLK MOSI MISO SCLK MOSI MISO SPI Slave Slave 1 Master 883 SCLK MOS MISO SPI Stave Slave 2 SCK Serial Clock MOSI Master Out Slave in SCLK MOS MISO MISO Master in Slave Out SPI Slave 3 Slave Slave Select
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Compute the executing time in microinstruction cycles for STOD, STOL, JPOS, JUMP, SUBL. Indicate which instructions are for Fetch, which are for Decode and which are for Execute.
Consider the following MARIE Assembly program: label opcode operand hex 100 JNS func 0106 101 LOAD X 1104 102 OUTPUT 6000 103 HALT 7000 104 DEC 5 0005 105 DEC 1 0001 106 DEC 0 0000 107 LOAD X 1104 108 ADD Y 3105 109 STORE 2104 10A JUMPI func C106 Copyright 2016-2021 by A.Berrached-- All Rights Reserved. What's stored in the PC (Program Counter) register right after the JNS instruction is executed? Express in Hex. Question 32 Consider the following MARIE O X Y func 27
MOD=11 EFFECTIVE ADDRESS CALCULATION MOD= 00 R/M W=0 W=1 R/M MOD =01 MOD = 10 00 AL AX 000 (BX) + (SI) (BX) + (DI) (BP) + (SI) (BP) + (DI) (SI) (DI) DIRECT ADDRESS (BX) (BX) + (SI) + D8 (BX) + (DI) + D8 (BP) + (SI) + D8 (BP) + (DI) + D8 (S) + D8 (DI) + D8 (BP) + D8 (BX) + D8 (BX) + (SI) + D16 (BX) + (DI) + D16 (BP) + (SI) + D16 (BP) + (DI) + D16 (SI) + D16 (DI) + D16 (BP) + D16 (BX) + D16 001 CL CX DX 001 010 DL 010 011 011 BL BX 100 101 AH SP 100 CH BP 101 110 DH SI 110 111 BH DI 111 Figure: R/M vs MOD Chart for MOV: 100010 instruction REGISTER SELECT (SEE FIGURE 3-7) BYTE 1 BYTE 3 BYTE 4 1000 10| OP CODE LOW DISPLACEMENT HIGH DISPLACEMENT DW MOD REG RM OR DIRECT ADDRESS DIREST ADDRESS HIGH BYTE LOW BYTE (5 BITS) ADDRESSING MODE (SEE FIGURE 3.8) BYTEWORD DATA O- BYTE 1- WORD DIRECTION TOFROM REG 0= FROM 1= TO OPERATION CODE Figure: MOV: 100010 instruction template
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A B I, 0,0 I,A,1 I, 1,1 I, 2,B I, 1,3 I,3,2 I, 2,4 RR,C I, 3,4 RNR, 4 RR,4,P RNR, 4, D RR,E,P RR, 4, F 1,4,4
Bus d'adresse 0x0 0 A7 Oxo 0 A6 0 A5 0 A4 0 A3 0 A2 0 A1 0 AO Bus de contrôle Écriture Lecture Bus de données Mémoire d'Instructions Adr Adr Ctr Ctr Ctr Ox Mémoire de données Clavier Écran 0 EN Data Data ΕΝ Data EN Data EN AO A1 DO D1 DECONCUL D3 D2 Write the instruction JZE R2, #OxOF in hexadecimal (using 2 bytes). a (Ctrl)
Assume that core components of single-cycle processor (shown below) have the following latencies: Regs Rd/Wr 80ps/60ps Adder Sign-Extend 20ps I-Mem Mux ALU D-Mem Shift-Left-2 40ps 50ps 20ps 100ps 200ps 20ps Single-cycle processor data path: Left [25-0 [31-26] PC+4 [31-28 Shift Let [25-21] Read Reg. 1 CLK - (20-16] Read Reg. 2 Addr. Data Read 15-11 Write Zero data -Cache / -MEM Res. Addr. Read Wite Data data 2 Read Data Register File Write Data RegWrite RegDst Branch (15-0) Sign Extend D-Cache D-Mem ALU 32 control ALUSrc Control InstType Memwrite MemtoReg MemRead Jump Suppose that this data path executes only two types of instruction: sub $rd, $rs, $rt lw $rt, offset ($rs) What would be the clock period to correctly execute the two instructions on the above single-cycle processor? Assume that PC register doesn't take any latency (i.e. Propagating a new PC value to I-Cache/I-Mem doesn't take any cycle).
Describe addl moub subq incw the following addressing modes. 12 (%rbp), % ecx (%rax, rcx), % de %rdx, (%rcx, %orax, 8) OXA (%rcx, 8)
14 Assume that the base address of 8255-PPI chip is OC00H and the address of port C of the chip is OC04H, which address lines are generating the selection line A1 A0? * Selection lines A1 A0 from address lines A2 A1 O None Selection lines A1 A0 from address lines A1 A0 Selection lines A1 A0 from address lines A3 A2 O Selection lines A1 A0 from address lines A2 A0
EAX=0019FFCC EBX-003AA000 ECX-00401005 ESI=00401005 EIP=00401015 EDX=00401005 EDI=00401005 EBP=0019FF80 ESP=0019FF74 EFL=00000246 CF=0 SF=0 ZF=1 OF=0 A What is DX? B What is the decimal value of EAX (Leave as the sum of POWERS) C What is EDX in binary? D What is DL AND CL? E What is DL OR CL? What is the value of Dec ESI G If signed, eax is now FFCC, Is it + or -? H How can you print out its correct decimal value if it is negative I Using shr and AND change ebx to AA00

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O LDA 8 1 LDB 9 2 INP 8 3 INP 9 4 ADD 5 STR 10 6 PNT 10 7 STP 8 10 9 20 10